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Difference between revisions of "2019 AIME II Problems/Problem 9"

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Why are you on this page? You should be preparing for this contest, not wasting your time looking for the solutions!
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==Problem==
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Call a positive integer <math>n</math> <math>k</math>-<i>pretty</i> if <math>n</math> has exactly <math>k</math> positive divisors and <math>n</math> is divisible by <math>k</math>. For example, <math>18</math> is <math>6</math>-pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\tfrac{S}{20}</math>.
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==Solution 1==
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Every 20-pretty integer can be written in form <math>n = 2^a 5^b k</math>, where <math>a \ge 2</math>, <math>b \ge 1</math>, <math>\gcd(k,10) = 1</math>, and <math>d(n) = 20</math>, where <math>d(n)</math> is the number of divisors of <math>n</math>. Thus, we have <math>20 = (a+1)(b+1)d(k)</math>, using the fact that the divisor function is multiplicative. As <math>(a+1)(b+1)</math> must be a divisor of 20, there are not many cases to check.
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If <math>a+1 = 4</math>, then <math>b+1 = 5</math>. But this leads to no solutions, as <math>(a,b) = (3,4)</math> gives <math>2^3 5^4 > 2019</math>.
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If <math>a+1 = 5</math>, then <math>b+1 = 2</math> or <math>4</math>. The first case gives <math>n = 2^4 \cdot 5^1 \cdot p</math> where <math>p</math> is a prime other than 2 or 5. Thus we have <math>80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23</math>. The sum of all such <math>n</math> is <math>80(3+7+11+13+17+19+23) = 7440</math>. In the second case <math>b+1 = 4</math> and <math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>.
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If <math>a+1 = 10</math>, then <math>b+1 = 2</math>, but this gives <math>2^9 \cdot 5^1 > 2019</math>. No other values for <math>a+1</math> work.
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Then we have <math>\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}</math>.
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-scrabbler94
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==Solution 2==
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For <math>n</math> to have exactly <math>20</math> positive divisors, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math> where <math>p,q,r</math> are primes. No number that is a multiple of <math>20</math> can be expressed in the first form because 20 has ''two'' primes in its prime factorization, while the first form has only ''one'', and the only integer divisible by <math>20</math> that has the second form is <math>2^{9}5</math>, which is 2560, greater than <math>2019</math>.
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For the third form, the only <math>20</math>-pretty numbers are <math>2^45^3=2000</math> and <math>2^35^4=5000</math>, and only <math>2000</math> is small enough.
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For the fourth form, any number of the form <math>2^45r</math> where <math>r</math> is a prime other than <math>2</math> or <math>5</math> will satisfy the <math>20</math>-pretty requirement. Since <math>n=80r<2019</math>, <math>r\le 25</math>. Therefore, <math>r</math> can take on <math>3, 7, 11, 13, 17, 19,</math> or <math>23</math>.
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Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>.
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Rephrased for clarity by Afly
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==See Also==
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{{AIME box|year=2019|n=II|num-b=8|num-a=10}}
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[[Category: Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 00:38, 22 December 2024

Problem

Call a positive integer $n$ $k$-pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$-pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$-pretty. Find $\tfrac{S}{20}$.

Solution 1

Every 20-pretty integer can be written in form $n = 2^a 5^b k$, where $a \ge 2$, $b \ge 1$, $\gcd(k,10) = 1$, and $d(n) = 20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20 = (a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.

If $a+1 = 4$, then $b+1 = 5$. But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$.

If $a+1 = 5$, then $b+1 = 2$ or $4$. The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$. The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$. In the second case $b+1 = 4$ and $d(k) = 1$, and there is one solution $n = 2^4 \cdot 5^3 = 2000$.

If $a+1 = 10$, then $b+1 = 2$, but this gives $2^9 \cdot 5^1 > 2019$. No other values for $a+1$ work.

Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$.

-scrabbler94

Solution 2

For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$ where $p,q,r$ are primes. No number that is a multiple of $20$ can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has only one, and the only integer divisible by $20$ that has the second form is $2^{9}5$, which is 2560, greater than $2019$.

For the third form, the only $20$-pretty numbers are $2^45^3=2000$ and $2^35^4=5000$, and only $2000$ is small enough.

For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$-pretty requirement. Since $n=80r<2019$, $r\le 25$. Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$.

Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$.

Rephrased for clarity by Afly

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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